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2024-04-24 19:07:14
Ta có: \(6xy+4x+15y+18=0\\ \Leftrightarrow\left(6xy+15y\right)+\left(4x+18\right)=0\\ \Leftrightarrow3y\left(2x+5\right)+2\left(2x+6\right)=0\Leftrightarrow3y\left(2x+5\right)+2\left(2x+5\right)+2=0\\ \Leftrightarrow\left(2x+5\right)\left(3y+2\right)=-2\)
Vì \(x,y\inℤ\) nên \(2x+5\inℤ;3y+2\inℤ\)
\(\Rightarrow\left(2x+5;3y+2\right)\inƯ\left(-2\right)\\\Rightarrow\left(2x+5;3y+2\right)\in \left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
| \(2x+5\) | \(1\) | \(2\) | \(-1\) | \(-2\) |
| \(3y+2\) | \(2\) | \(1\) | \(-2\) | \(-1\) |
| \(x\) | \(\dfrac{8}{11}\) | \(\dfrac{1}{11}\) | \(-\dfrac{8}{11}\) | \(-\dfrac{1}{11}\) |
| \(y\) | \(-\dfrac{1}{11}\) | \(\dfrac{4}{11}\) | \(\dfrac{1}{11}\) | \(-\dfrac{4}{11}\) |
\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{8}{11};-\dfrac{1}{11}\right);\left(\dfrac{1}{11};\dfrac{4}{11}\right);\left(-\dfrac{8}{11};\dfrac{1}{11}\right);\left(-\dfrac{1}{11};-\dfrac{4}{11}\right)\right\}\)(Loại)
Vậy không có nghiệm \(x,y\inℤ\)